Show that $infleft{b/a:bin B,ain Aright}=inf{B}/sup{A}$
$begingroup$
Let $ A $, $ B $ two subsets of the real line; let, for at least one $ a>0 $, $ ain A $. I've tried to prove that $ infleft(B/A_{>0}right)=inf{B}/sup{A} $, were $ A_{>0} $ is meant to be $ mathbb{R}_{>0}cap A $. The thesis sounds "geometrically" evident to me.
My poor attempt of proof: It's clear that $ inf{B}/sup{A} $ should be a lower bound for $ B/A_{>0} $. The first thing that came to my mind to show that, given $ x>inf{B}/sup{A} $, there exists $ b'/a'in B/A_{>0} $ such that $ b'/a'<x $, was to consider $ epsilon>0 $ then let $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $. There exists consequently $ b'<inf{B}epsilon $ and $ a'>sup{A}(1/epsilon) $ and $$ frac{b'}{a'}<frac{inf{B}}{sup{A}}cdotepsilon^2 $$ actually proving the statement.
This approach is obviously wrong, since $ x $ and $ y $ need not to be respectively greater than $ inf{B} $ and lower than $ sup{A} $.
Is this statement at all true? If so, may I have a hint or a proof?
real-analysis supremum-and-infimum
asked Dec 16 '18 at 17:21
marco21marco21
308211
$endgroup$
add a comment |
$begingroup$
Let $ A $, $ B $ two subsets of the real line; let, for at least one $ a>0 $, $ ain A $. I've tried to prove that $ infleft(B/A_{>0}right)=inf{B}/sup{A} $, were $ A_{>0} $ is meant to be $ mathbb{R}_{>0}cap A $. The thesis sounds "geometrically" evident to me.
My poor attempt of proof: It's clear that $ inf{B}/sup{A} $ should be a lower bound for $ B/A_{>0} $. The first thing that came to my mind to show that, given $ x>inf{B}/sup{A} $, there exists $ b'/a'in B/A_{>0} $ such that $ b'/a'<x $, was to consider $ epsilon>0 $ then let $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $. There exists consequently $ b'<inf{B}epsilon $ and $ a'>sup{A}(1/epsilon) $ and $$ frac{b'}{a'}<frac{inf{B}}{sup{A}}cdotepsilon^2 $$ actually proving the statement.
This approach is obviously wrong, since $ x $ and $ y $ need not to be respectively greater than $ inf{B} $ and lower than $ sup{A} $.
Is this statement at all true? If so, may I have a hint or a proof?
real-analysis supremum-and-infimum
asked Dec 16 '18 at 17:21
marco21marco21
308211
$endgroup$
1
$begingroup$
I think that the statement is true only if $B$ consists of positive numbers.
$endgroup$
– N. S.
Dec 16 '18 at 17:45
$begingroup$
@N.S. That's true. But how to show it anyway? Considering $ epsilon > 0 $ and taking $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $ does not necessarily give $ x>inf{B} $ and $ y<sup{A} $, so my proof is wrong even adding this constraint to my hypothesis.
$endgroup$
– marco21
Dec 16 '18 at 18:02
1
$begingroup$
If $x>inf{B}/sup{A}$ then $x sup(A)> inf(B)$. Then, there exists some $b in B$ such that $x sup(A) > b$. Now, and here is you positivity, since $x$ is positive (note also that $x$ cannot be zero) you get $$sup(A)> frac{b}{x} ,$$ Thus you can find some $a in A$ such that $a>frac{b}{x}$.
$endgroup$
– N. S.
Dec 16 '18 at 22:29
$begingroup$
Thank you very much for the clarification!
$endgroup$
– marco21
Dec 16 '18 at 23:51
$begingroup$
np... Note that your problem can be rewritten in the form $$infleft(B/A_{>0}right) cdot sup{A} =inf{B}$$ which is easier to deal with, because there is no fraction. I used this trick in the solution.
$endgroup$
– N. S.
Dec 17 '18 at 1:58
add a comment |
$begingroup$
Let $ A $, $ B $ two subsets of the real line; let, for at least one $ a>0 $, $ ain A $. I've tried to prove that $ infleft(B/A_{>0}right)=inf{B}/sup{A} $, were $ A_{>0} $ is meant to be $ mathbb{R}_{>0}cap A $. The thesis sounds "geometrically" evident to me.
My poor attempt of proof: It's clear that $ inf{B}/sup{A} $ should be a lower bound for $ B/A_{>0} $. The first thing that came to my mind to show that, given $ x>inf{B}/sup{A} $, there exists $ b'/a'in B/A_{>0} $ such that $ b'/a'<x $, was to consider $ epsilon>0 $ then let $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $. There exists consequently $ b'<inf{B}epsilon $ and $ a'>sup{A}(1/epsilon) $ and $$ frac{b'}{a'}<frac{inf{B}}{sup{A}}cdotepsilon^2 $$ actually proving the statement.
This approach is obviously wrong, since $ x $ and $ y $ need not to be respectively greater than $ inf{B} $ and lower than $ sup{A} $.
Is this statement at all true? If so, may I have a hint or a proof?
real-analysis supremum-and-infimum
asked Dec 16 '18 at 17:21
marco21marco21
308211
$endgroup$
Let $ A $, $ B $ two subsets of the real line; let, for at least one $ a>0 $, $ ain A $. I've tried to prove that $ infleft(B/A_{>0}right)=inf{B}/sup{A} $, were $ A_{>0} $ is meant to be $ mathbb{R}_{>0}cap A $. The thesis sounds "geometrically" evident to me.
My poor attempt of proof: It's clear that $ inf{B}/sup{A} $ should be a lower bound for $ B/A_{>0} $. The first thing that came to my mind to show that, given $ x>inf{B}/sup{A} $, there exists $ b'/a'in B/A_{>0} $ such that $ b'/a'<x $, was to consider $ epsilon>0 $ then let $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $. There exists consequently $ b'<inf{B}epsilon $ and $ a'>sup{A}(1/epsilon) $ and $$ frac{b'}{a'}<frac{inf{B}}{sup{A}}cdotepsilon^2 $$ actually proving the statement.
This approach is obviously wrong, since $ x $ and $ y $ need not to be respectively greater than $ inf{B} $ and lower than $ sup{A} $.
Is this statement at all true? If so, may I have a hint or a proof?
real-analysis supremum-and-infimum
real-analysis supremum-and-infimum
asked Dec 16 '18 at 17:21
marco21marco21
308211
asked Dec 16 '18 at 17:21
marco21marco21
308211
asked Dec 16 '18 at 17:21
marco21marco21
308211
asked Dec 16 '18 at 17:21
marco21marco21
308211
asked Dec 16 '18 at 17:21
marco21marco21
308211
308211
1
$begingroup$
I think that the statement is true only if $B$ consists of positive numbers.
$endgroup$
– N. S.
Dec 16 '18 at 17:45
$begingroup$
@N.S. That's true. But how to show it anyway? Considering $ epsilon > 0 $ and taking $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $ does not necessarily give $ x>inf{B} $ and $ y<sup{A} $, so my proof is wrong even adding this constraint to my hypothesis.
$endgroup$
– marco21
Dec 16 '18 at 18:02
1
$begingroup$
If $x>inf{B}/sup{A}$ then $x sup(A)> inf(B)$. Then, there exists some $b in B$ such that $x sup(A) > b$. Now, and here is you positivity, since $x$ is positive (note also that $x$ cannot be zero) you get $$sup(A)> frac{b}{x} ,$$ Thus you can find some $a in A$ such that $a>frac{b}{x}$.
$endgroup$
– N. S.
Dec 16 '18 at 22:29
$begingroup$
Thank you very much for the clarification!
$endgroup$
– marco21
Dec 16 '18 at 23:51
$begingroup$
np... Note that your problem can be rewritten in the form $$infleft(B/A_{>0}right) cdot sup{A} =inf{B}$$ which is easier to deal with, because there is no fraction. I used this trick in the solution.
$endgroup$
– N. S.
Dec 17 '18 at 1:58
add a comment |
1
$begingroup$
I think that the statement is true only if $B$ consists of positive numbers.
$endgroup$
– N. S.
Dec 16 '18 at 17:45
$begingroup$
@N.S. That's true. But how to show it anyway? Considering $ epsilon > 0 $ and taking $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $ does not necessarily give $ x>inf{B} $ and $ y<sup{A} $, so my proof is wrong even adding this constraint to my hypothesis.
$endgroup$
– marco21
Dec 16 '18 at 18:02
1
$begingroup$
If $x>inf{B}/sup{A}$ then $x sup(A)> inf(B)$. Then, there exists some $b in B$ such that $x sup(A) > b$. Now, and here is you positivity, since $x$ is positive (note also that $x$ cannot be zero) you get $$sup(A)> frac{b}{x} ,$$ Thus you can find some $a in A$ such that $a>frac{b}{x}$.
$endgroup$
– N. S.
Dec 16 '18 at 22:29
$begingroup$
Thank you very much for the clarification!
$endgroup$
– marco21
Dec 16 '18 at 23:51
$begingroup$
np... Note that your problem can be rewritten in the form $$infleft(B/A_{>0}right) cdot sup{A} =inf{B}$$ which is easier to deal with, because there is no fraction. I used this trick in the solution.
$endgroup$
– N. S.
Dec 17 '18 at 1:58
1
1
$begingroup$
I think that the statement is true only if $B$ consists of positive numbers.
$endgroup$
– N. S.
Dec 16 '18 at 17:45
$begingroup$
I think that the statement is true only if $B$ consists of positive numbers.
$endgroup$
– N. S.
Dec 16 '18 at 17:45
$begingroup$
@N.S. That's true. But how to show it anyway? Considering $ epsilon > 0 $ and taking $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $ does not necessarily give $ x>inf{B} $ and $ y<sup{A} $, so my proof is wrong even adding this constraint to my hypothesis.
$endgroup$
– marco21
Dec 16 '18 at 18:02
$begingroup$
@N.S. That's true. But how to show it anyway? Considering $ epsilon > 0 $ and taking $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $ does not necessarily give $ x>inf{B} $ and $ y<sup{A} $, so my proof is wrong even adding this constraint to my hypothesis.
$endgroup$
– marco21
Dec 16 '18 at 18:02
1
1
$begingroup$
If $x>inf{B}/sup{A}$ then $x sup(A)> inf(B)$. Then, there exists some $b in B$ such that $x sup(A) > b$. Now, and here is you positivity, since $x$ is positive (note also that $x$ cannot be zero) you get $$sup(A)> frac{b}{x} ,$$ Thus you can find some $a in A$ such that $a>frac{b}{x}$.
$endgroup$
– N. S.
Dec 16 '18 at 22:29
$begingroup$
If $x>inf{B}/sup{A}$ then $x sup(A)> inf(B)$. Then, there exists some $b in B$ such that $x sup(A) > b$. Now, and here is you positivity, since $x$ is positive (note also that $x$ cannot be zero) you get $$sup(A)> frac{b}{x} ,$$ Thus you can find some $a in A$ such that $a>frac{b}{x}$.
$endgroup$
– N. S.
Dec 16 '18 at 22:29
$begingroup$
Thank you very much for the clarification!
$endgroup$
– marco21
Dec 16 '18 at 23:51
$begingroup$
Thank you very much for the clarification!
$endgroup$
– marco21
Dec 16 '18 at 23:51
$begingroup$
np... Note that your problem can be rewritten in the form $$infleft(B/A_{>0}right) cdot sup{A} =inf{B}$$ which is easier to deal with, because there is no fraction. I used this trick in the solution.
$endgroup$
– N. S.
Dec 17 '18 at 1:58
$begingroup$
np... Note that your problem can be rewritten in the form $$infleft(B/A_{>0}right) cdot sup{A} =inf{B}$$ which is easier to deal with, because there is no fraction. I used this trick in the solution.
$endgroup$
– N. S.
Dec 17 '18 at 1:58
add a comment |
1 Answer
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$begingroup$
This is not true. If $B=[-1,0]$ and $A=(0,1]$, $inf B/sup A= -1$ even though $inf (B/A)=-infty$. If we assume $Bsubset (0,infty)$, then $inf B/sup Ale b/sup Ale b/a$ for $ain A_{>0}$ so $inf B/sup Ale inf (B/A_{>0})$. Since $ainf (B/A_{>0})le b$ for all $ain A_{>0}$, $ainf (B/A_{>0})le inf B$. If $inf (B/A_{>0})=0$ then $inf B=0$ or $sup A=infty$ and we are done (assuming $b/infty=0$). Otherwise, $sup Ale inf B/inf(B/A_{>0})$, which gives the required reverse inequality $inf(B/A_{>0})=inf B/sup A$.
answered Dec 16 '18 at 17:43
Guacho PerezGuacho Perez
3,93411133
$endgroup$
add a comment |
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$begingroup$
This is not true. If $B=[-1,0]$ and $A=(0,1]$, $inf B/sup A= -1$ even though $inf (B/A)=-infty$. If we assume $Bsubset (0,infty)$, then $inf B/sup Ale b/sup Ale b/a$ for $ain A_{>0}$ so $inf B/sup Ale inf (B/A_{>0})$. Since $ainf (B/A_{>0})le b$ for all $ain A_{>0}$, $ainf (B/A_{>0})le inf B$. If $inf (B/A_{>0})=0$ then $inf B=0$ or $sup A=infty$ and we are done (assuming $b/infty=0$). Otherwise, $sup Ale inf B/inf(B/A_{>0})$, which gives the required reverse inequality $inf(B/A_{>0})=inf B/sup A$.
answered Dec 16 '18 at 17:43
Guacho PerezGuacho Perez
3,93411133
$endgroup$
add a comment |
$begingroup$
This is not true. If $B=[-1,0]$ and $A=(0,1]$, $inf B/sup A= -1$ even though $inf (B/A)=-infty$. If we assume $Bsubset (0,infty)$, then $inf B/sup Ale b/sup Ale b/a$ for $ain A_{>0}$ so $inf B/sup Ale inf (B/A_{>0})$. Since $ainf (B/A_{>0})le b$ for all $ain A_{>0}$, $ainf (B/A_{>0})le inf B$. If $inf (B/A_{>0})=0$ then $inf B=0$ or $sup A=infty$ and we are done (assuming $b/infty=0$). Otherwise, $sup Ale inf B/inf(B/A_{>0})$, which gives the required reverse inequality $inf(B/A_{>0})=inf B/sup A$.
answered Dec 16 '18 at 17:43
Guacho PerezGuacho Perez
3,93411133
$endgroup$
add a comment |
$begingroup$
This is not true. If $B=[-1,0]$ and $A=(0,1]$, $inf B/sup A= -1$ even though $inf (B/A)=-infty$. If we assume $Bsubset (0,infty)$, then $inf B/sup Ale b/sup Ale b/a$ for $ain A_{>0}$ so $inf B/sup Ale inf (B/A_{>0})$. Since $ainf (B/A_{>0})le b$ for all $ain A_{>0}$, $ainf (B/A_{>0})le inf B$. If $inf (B/A_{>0})=0$ then $inf B=0$ or $sup A=infty$ and we are done (assuming $b/infty=0$). Otherwise, $sup Ale inf B/inf(B/A_{>0})$, which gives the required reverse inequality $inf(B/A_{>0})=inf B/sup A$.
answered Dec 16 '18 at 17:43
Guacho PerezGuacho Perez
3,93411133
$endgroup$
This is not true. If $B=[-1,0]$ and $A=(0,1]$, $inf B/sup A= -1$ even though $inf (B/A)=-infty$. If we assume $Bsubset (0,infty)$, then $inf B/sup Ale b/sup Ale b/a$ for $ain A_{>0}$ so $inf B/sup Ale inf (B/A_{>0})$. Since $ainf (B/A_{>0})le b$ for all $ain A_{>0}$, $ainf (B/A_{>0})le inf B$. If $inf (B/A_{>0})=0$ then $inf B=0$ or $sup A=infty$ and we are done (assuming $b/infty=0$). Otherwise, $sup Ale inf B/inf(B/A_{>0})$, which gives the required reverse inequality $inf(B/A_{>0})=inf B/sup A$.
answered Dec 16 '18 at 17:43
Guacho PerezGuacho Perez
3,93411133
edited Dec 16 '18 at 18:04
answered Dec 16 '18 at 17:43
Guacho PerezGuacho Perez
3,93411133
answered Dec 16 '18 at 17:43
Guacho PerezGuacho Perez
3,93411133
answered Dec 16 '18 at 17:43
Guacho PerezGuacho Perez
3,93411133
3,93411133
add a comment |
add a comment |
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$begingroup$
I think that the statement is true only if $B$ consists of positive numbers.
$endgroup$
– N. S.
Dec 16 '18 at 17:45
$begingroup$
@N.S. That's true. But how to show it anyway? Considering $ epsilon > 0 $ and taking $ x=inf{B}epsilon $ and $ y=sup{A}(1/epsilon) $ does not necessarily give $ x>inf{B} $ and $ y<sup{A} $, so my proof is wrong even adding this constraint to my hypothesis.
$endgroup$
– marco21
Dec 16 '18 at 18:02
1
$begingroup$
If $x>inf{B}/sup{A}$ then $x sup(A)> inf(B)$. Then, there exists some $b in B$ such that $x sup(A) > b$. Now, and here is you positivity, since $x$ is positive (note also that $x$ cannot be zero) you get $$sup(A)> frac{b}{x} ,$$ Thus you can find some $a in A$ such that $a>frac{b}{x}$.
$endgroup$
– N. S.
Dec 16 '18 at 22:29
$begingroup$
Thank you very much for the clarification!
$endgroup$
– marco21
Dec 16 '18 at 23:51
$begingroup$
np... Note that your problem can be rewritten in the form $$infleft(B/A_{>0}right) cdot sup{A} =inf{B}$$ which is easier to deal with, because there is no fraction. I used this trick in the solution.
$endgroup$
– N. S.
Dec 17 '18 at 1:58